Superposition Method
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1 C4303 Structural Theory 1 Superposition Method En Hamizi Yahya & Pn Nasirah Siron Reference: Tn Hj Ahmad Shufni Analyzing can be done at cantilever beam that is at support at free end section The beam above is carrying uniformly distributed load (UDL) that may caused reaction acting up at free end, B By removing the support at B, deflection y B is occurred at end B P By placing support at end B equivalent to reaction force P acting up. (equivalent to reaction due to support B) yB A B A B A B A B
description
CC505 Structure Analysis 1
Transcript of Superposition Method
1
C4303 Structural Theory 1
Superposition Method
En Hamizi Yahya & Pn Nasirah Siron
Reference: Tn Hj Ahmad Shufni
Analyzing can be done at cantilever beam that is at support at free end section
The beam above is carrying uniformly distributed load (UDL) that may caused reaction acting up at free end, B
By removing the support at B, deflection yB is occurred at end B
P
By placing support at end B equivalent to reaction force P acting up. (equivalent to reaction due to support B)
yB
A B
A B
A B
A B
2
By equating the deflection acting down yw due to loading and up due to reaction forced P (yp) the reaction
force P can be determined
OR to prevent deflection, deflection at end B MUST in zero deflection
So yB + yP = 0
Problem 1
By superposition method, determine the reaction at support B,
a) The reaction at support A and sketch Shear Force Diagram (SFD)
b) Bending moment at A and B
c) Sketch the Bending Moment Diagram (BMD)
10 kN
2m 2m
Solution
Separate beam into 2 condition:
Case 1 : Take away support at B in order to transform simply supported into cantilever beam
Case 2 : Move all the loading at beam and place the reaction for P at B
Support at B is moved away so it will be cantilever beam. Due to loading (Point load P and UDL),
deflection occurred at free end at B yw.
Derive moment equation, at section z-z where as near to fixed end A. Moment distance (x) is taken from
free end.
Case 1
10 kN
z
Mx yw
z x
Moment equation
Mx + 10 (x-2) = 0
Mx = EId2y
dx2
EId2y
dx 2 = −10(x − 2)---------------------------- eq. 1
A B
A B
3
Integrate moment equation 1 toward x, to obtain slope deflection equation.
= EIdy
dx=
−10(x−2)2
2+ A -----------------------eq. 2
Integrate slope deflection equation 2 toward x, to obtain deflection equation.
= EIyw = −10(x−2)3
6+ Ax + B--------------------eq. 3
To state the boundary condition :
At A when x = 4 m, slope = 0 from equation 2
0 = −10(4 − 2)2
2+ A
A = 20
At A when x = 4 m, deflection = 0
From equation 3
0 = −10(4−2)3
6+ 20(4) + B
B = -66.67
At B when x = 0, deflection yw = ??
From equation 3
EIyw = −10(0−2)3
6+ 20 0 − 66.67
yw = −66.67
𝐸𝐼
Case 2
Move all the loading at cantilever beam, and at free end B, put the reaction at free end B, P
z
yP
Mx
z x P
A B
Derive moment equation toward section z-z where as near to fixed-end, A. Moment distance, x is taken
from free end
Mx – Px = 0
Mx = EId2y
dx2
EId2y
dx 2 = Px-----------------------eq. 1
Integrate moment equation 1 toward x, to obtain slope deflection equation.
0 0
A B
4
= EIdy
dx=
Px 2
2+ A -----------------------eq. 2
Integrate slope deflection equation 2 toward x, to obtain deflection equation.
= EIyw = Px
3
6+ Ax + B--------------------eq. 3
To state the boundary condition :
At A when x = 4 m, slope = 0 from equation 2
0 = P(4)2
2+ A
A = 8P
At A when x = 4 m, deflection = 0
From equation 3
EI(0) = P(4)3
6+ 8P(4) + B
B=21.33P
At B when x = 0, deflection yp = ??
From equation 3
EIyp = P(0)3
6+ 8P(0) + 21.33P
yp = 21.33P
EI
Super imposed case 1 and case 2, to obtain deflection = 0 (stability)
yw + yp = 0
−66.67
EI+
21.33P
EI= 0
P = 3.13 kN
5
10 kN
MA = -7.48 kNm
2m 2m
A B
RA= 6.87 kN RB = P = 3.13 kN
6.87kN 6.87kN
SFD
-7.48kNm -3.13kN -3.13kN
BMD 0kNm
6.26kNm
Problem 2
By superposition method, determine the reaction at support B,
a) The reaction at support A and sketch Shear Force Diagram (SFD)
b) Bending moment at A and B
c) Sketch the Bending Moment Diagram (BMD)
5 kN/m
4m
Fy = 0
Fy = Fy
RA + 3.13 = 10
RA = 6.87 kN
M@A = 0
MA + 10(2) – 3.13(4) = 0
MA = -7.48 kNm
A B
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Solution
Case1
Z 5 kN/m
Mx yw
z x
Moment equation
Mx + 5 𝑥 (𝑥)
2= 0
Mx = EId2y
dx2
EId2y
dx 2 = −5 𝑥
2
2
-------------------------- eq. 1
Integrate moment equation 1 toward x, to obtain slope deflection equation.
= EIdy
dx= −
5(x)3
6+ A -----------------------eq. 2
Integrate slope deflection equation 2 toward x, to obtain deflection equation.
= EIyw = − −5(x)4
24+ Ax + B--------------------eq. 3
To state the boundary condition :
At A when x = 4 m, slope = 0 from equation 2
EIdy
dx= −
5(4)3
6+ A
A = 53.33
At A when x = 4 m, deflection = 0 from equation 3
EI(0) = − −5(4)4
24+ 53.33(4) + B
B = -159.99
At B when x = 0, deflection yw = ??
From equation 3
EIyw = −5(0)4
24+ 53.33 0 − 159.99
yw = −159.99
𝐸𝐼
0 0
A B
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Case 2
Move all the loading at cantilever beam, and at free end B, put the reaction at free end B, P
z
yP
Mx
z x P
Derive moment equation toward section z-z where as near to fixed-end, A. Moment distance, x is taken
from free end
Mx – Px = 0
Mx = EId2y
dx2
EId2y
dx 2 = Px-----------------------eq. 1
Integrate moment equation 1 toward x, to obtain slope deflection equation.
= EIdy
dx=
Px 2
2+ A -----------------------eq. 2
Integrate slope deflection equation 2 toward x, to obtain deflection equation.
= EIyw = Px
3
6+ Ax + B--------------------eq. 3
To state the boundary condition :
At A when x = 4 m, slope = 0 from equation 2
0 = P(4)2
2+ A
A = 8P
At A when x = 4 m, deflection = 0
From equation 3
EI(0) = P(4)3
6+ 8P(4) + B
B=21.33P
At B when x = 0, deflection yp = ??
From equation 3
EIyp = P(0)3
6+ 8P(0) + 21.33P
yp = 21.33P
EI
A B
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Super imposed case 1 and case 2, to obtain deflection = 0 (stability)
yw + yp = 0
−159.99
EI+
21.33P
EI= 0
P = 7.5 kN
MA = -10 kNm
5 kN/m
4m
RA= 12.5 kN RB = P = 7.5 kN
12.5kN
SFD
2.5m 1.5m
-7.5kN
-10kNm
BMD 0kNm
5.63kNm
Fy = 0
Fy = Fy
RA + 7.5 = 5(4)
RA = 12.5 kN
M@A = 0
MA + 5(4)(4/2) – 7.5(4) = 0
MA = -10 kNm
A B
